Show that the relations $\displaystyle (A \cup C) \subset (A \cup B) $ and $\displaystyle (A \cap C) \subset (A \cap B) $, when considered together, imply $\displaystyle C \subset B $.

Here is my proof:

If $\displaystyle x \in (A \cup C)$ then $\displaystyle x \in (A \cup B)$.

If $\displaystyle x \in (A \cup C)$ then $\displaystyle x \in A $ or $\displaystyle x \in C$.

We want to know about the elements in C so we consider $\displaystyle x \in C$.

If $\displaystyle x \in C$ then $\displaystyle x \in (A \cup B)$.

This means that if $\displaystyle x \in C$ then $\displaystyle x \in A$, $\displaystyle x \in B$, or $\displaystyle x \in A$ and $\displaystyle x \in B$.

There are three cases.

First: $\displaystyle x \in C, x \in A \implies x \in (A \cap C)$. From the second relation, $\displaystyle x \in (A \cap B)$, and therefore $\displaystyle x \in B$.

Second: $\displaystyle x \in C, x \in B \implies x \in B$.

Third: $\displaystyle x \in C, x \in (A \cap B) \implies x \in B$.

Thus $\displaystyle x \in C \implies x \in B$, therefore $\displaystyle C \subset B$.

Is this proof complete, correct, clean, nasty?

Any help is appreciated, especially cleaning it up and making it look all nice with proper notation. Again, if it's wrong it would be more helpful if you could point out where I've went astray.

Thanks,

ultros